Question

Find the middle term of the sequence formed by all three digit numbers which leave a reminder three , when divided by 4 .also find the sum of all numbers on both sides of middle term separately.

Answer 1

The smallest and the greatest 3 digit numbers which give remainder 3 when divided by 4 are 103 and 999 respectively
A.P = 103, 107.........999.
an= a + (n-1) d
d= 107-103= 4
999= 103 + (n-1) 4
n= 225 
middle term= (n+1/2)th term 
225+1/2 = 113th term
a113= a +  112 d
a113= 103+ 112.4
a113=551..................................................1
1. Sum of numbers before middle term 
sum a1 to a112
Sn= n/2(2a+(n-1)d)
S112= 112/2(2.103 = (112-1)4)
S112= 36400 
Sum of all numbers from a1 to a225
S225= 225/2(2.103+(225-1)4)
S225=123975...................................................2

Now sum of all numbers after middle term 
a1114 to a225 
S225- S112= S113 to S225
123975 -36400 = S113 to S225
Sum of numbers from a113 to a225 is 87575 

We need to get the sum of numbers after middle term 
Therefore 
(Sum of numbers from a113 to a225) - a113= 87575- 551                (from 1)
Sum of numbers after the middle terrm =87024